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3 years ago

If a driver is tired, the thinking distance will be less. True or false.why?

Physics

1 answer:

Evgesh-ka [11]3 years ago

8 0

Answer:

False

Explanation:

When a driver is impaired the brain will need an extra second or two to wake up, after it does wake up it tends to drift off regularly.

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Which of the following lies in the ecliptic plane?

babymother [125]

<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>

The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>

<u />

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about $23\°$ approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

7 0

4 years ago

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

3 years ago

Help me with this physics math?

solong [7]

Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes

(6.4)^2 x 10^12

= 40.96 x 10^12 .

Now it's just a matter of mashing out the fraction.

The 'mantissa' (the number part) is

6/40.96 = 0.1465

and the order of magnitude is

10^24 / 10^12 = 10^12 .

Put it all together and you've got

1.465 x 10^11 .

4 0

3 years ago

A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was

AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

7 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago